Butterfly problem ?

Forum|Forum|3 years ago
December 31, 2022
2 replies
5890 views

Hello, Everyone.
From the following :
Given :
Let ABC be a triangle with incircle ( I ), and let ( I ) touch BC, CA, AB at A', B', C', respectively.
Drop perpendicular from A' onto B'C' at A''. Lines BA'' and CA meet at B''. Lines CA'' and AB meet
at C''.
Prove :
C'C'' + B'B'' = C''B''

( dark-green + green = red )

Thanks in advance for your time and help.
Best Regards.

Loi.

Math Homework

Best answer by ttokoro

For my above picture, at first point P on the circle (1) is selected between circular arc c'Pb'.

Then make tangent line c"Pb" at point P on the circle. This is the red line.

Then

Make △OB"B'≡△OB"P and make △OC"C'≡△OC"P.

Length of OB'=OC'=OP=radius r of the circle.

Angle of ∠OB'B"=∠OPB"=∠OC'C"=∠OPC"=90 deg. (Because the tangent line must be perpendicular to radius r.)

So

If we can show the intersection point M of line cc" and bb" is also on the line c'b', it is solved.

I

idahoan

12-Amethyst

Please attach a MathCAD worksheet that shows the problem you are experiencing. Then perhaps we can assist.

W

Werner_E

25-Diamond I

Actually it does not matter where you choose the point A'' on the line B'C' !

B''C'' will always be a tangent to the circle and so the two green lines will add up to the red one.

L

lvl107

Author

20-Turquoise

Many thanks for your time and hints, idahoan and Werner. @Werner_E , your mark, the above, is a more general statement !

And the butterfly problem should be stated as the following:

Let ABC be a triangle with incircle ( I ), and let ( I ) touch AC, AB at B', C', respectively. M is a
point on segment B'C'. Lines BM and CA meet at B''. Lines CM and AB meet at C''.
Prove :
C'C'' + B'B'' = C''B''.